Question

A machine gun is mounted on the top of tower 100m high.At what angle should the gun be inclined to cover a max range of firing the ground below?Muzzle speed=150m/s,take g=10m/s^2

Answer 1

Time is 4.51 s from what i've got if im correct

Answer 2

Working with rough approximations..if theta=45,Rmax=2250m,it should be nearby thatonly..iguess

Answer 3

@experimentX

Answer 4

let it be inclined at an angle \( \theta \), the flight time is calculate the total flight time.

Answer 5

\[ - 2 v \sin \theta/10 + 100/(v \sin \theta)\]

Answer 6

the horizontal range is given by
\[ f(\theta) = v \cos \theta * (- 2 v \sin \theta/10 + 100/(v \sin \theta)) \]
maximize this function for theta

Answer 7

* ... -2v is +2v sin ..

Answer 8

Corrcect answer (is different)
Denote \[v = v_0*Sin \theta\] Then flight time is (assuming that g = 10)\[(-v + \sqrt{20H + v^2})/10 + 2v/10 = 0.1(v + \sqrt{20H + v^2})\]
2v/10

Answer 9

Now during this time the projectile will cover the distance
\[V_0*Cos \theta * 0.1(v+\sqrt{20H + v^2}) = D(\theta,V_0)\]

Answer 10

http://www.wolframalpha.com/input/?i=maximize+150+cos%28x%29+*+%282+*+150+sin%28x%29%2F10+%2B+100%2F%28150+sin%28x%29%29%29
http://www.wolframalpha.com/input/?i=change+0.762081+radian+into+degrees

Answer 11

So You will have to differentiate with respect to theta the following expression
\[V_0 Cos \theta* \left( V_0 Sin \theta + \sqrt{H^2 + V_0^2 Sin^2 \theta}\right)\]

Answer 12

Did this question require so many efforts lol :/

Answer 13

The solution I provided is the most straightforward - start differentiating it

Answer 14

im not taught that much high standard atm

Answer 15

seems 43.664 is the ans

Answer 16

No "king's road" to correct answer - either sweat or nothing

Answer 17

Wolfram will A) will do him NO GODD on exam, B) will leave him WITHOUT KNOWLEDGE HOW TO SIMILAR PROBLEMS IN THE FUTURE

Answer 18

this is just calculus ... maximizing and minimizing. let me see if i can find any other method.

Answer 19

NO GOOD

Answer 20

Lets see indeed ..... .... VERY interesting indeed ..................

Answer 21

x=u^2sin2theta/g +100cot theta

Answer 22

Here,
u = 150 ms-1
And, Taking g = 10 m-2 (approximately)
Let 'θ' be the angle at which the machine gun should fire in order to cover a maximum distance.

Then , the horizontal component of velocity = 150 cos θ

And, the vertical component of velocity = 150 sin θ

If 'T' is the time of flight, then ,
Horizontal range, R = (150 cos θ ) x T .

The gun is mounted at the top of a tower 100 meters high .

Let us regard the positive direction of the position-axis as to be along the line from the top of tower in downward direction.

For motion along vertical :


Initial Velocity = -150 sin θ;

Distance covered = + 100 m

And, acceleration = + 10 ms-2

In time 'T' , the machine gun shot will reach maximum height and then reach the ground.

Now,

S = ut + 1/2 at2

Therefore, +100 = ( -150 sin θ ) T + 1/2 x T2

Or, T2 - ( 30 sin θ )T - 20 = 0

Therefore, T = - ( - 30 sin θ ) ± { ( - 30 sin θ)2 - 4 x 1 x (-20) }1/2/2

= 30 sin θ ± (900 sin2 θ + 80 )1/2/2

Or, T = 15 sinθ ± (225 sin2θ + 20)1/2


Now, range will be maximum, if time of flight is maximum .

Therefore, choosing positive sign ,we have

T = 15 sin θ + ( 225 sin2θ+ 20 )1/2

Hence, horizontal range covered,
R = 150 cos θ {15 sin θ+ ( 225 sin2θ + 20 )}1/2

The horizontal range is maximum, when θ = 45o.

But in the present case , the machine gun is mounted at height of 100 m .

Therefore, R will not be maximum for θ = 45o.It will be maximum for some value of θ close to 45o.

If we calculate values of R by setting θ = 43o, 43.5o, 44o, 45o, 46o and 47o, the values of R come out to be 2347 m, 2347.7 m, 2348 m, 2346 m, 2341 m and 2334 m respectively.

Thus R is maximum for value of θ some where between 43.5o and 44o.

Therefore , the mean value of θ = (43.5o + 44o)/2 = 43.75o.

The gun should be inclined at 43.75o to cover a maximum range of firing on the ground below

Read more: http://wiki.answers.com/Q/A_machine_gun_is_mounted_on_the_top_of_tower_100m_high_at_what_angle_should_the_gun_be_inclined_to_cover_a_max._range_of_firing_on_ground_below_The_muzzle_speed_of_bullet_is_150_meters_p_sec#ixzz26X9WBUbQ

Answer 23

IReadThat

Answer 24

\[ f(\theta) = v \cos \theta * (+2 v \sin \theta/10 + 100/(v \sin \theta)) \\
v^2 \sin 2\theta /10 + 100 \cot \theta \]
doesn't seem easy ... probably you have to resort to calculus.

Answer 25

THIS STATEMENT IS NOT TRUE
"Now, range will be maximum, if time of flight is maximum ."

Answer 26

@experimentX AS i have said above

Answer 27

guys
Ive..got...a...simple..solution..here...just..exaplain..me!

Answer 28

I wanted to write g - but @experimentX here spoiled us by writing 10 everywhere, so I followed his bad example. And in differentiating I just dropped 0.1 (1/g) because it does not change the solution